Question

given m∠aeb = 45°. ∠aec is a right angle. prove: eb bisects ∠aec. proof. we are given that m∠aeb = 45° and ∠aec is a right angle. the measure of ∠aec is 90° by the definition of a right angle. applying the segment addition property gives m∠aeb + m∠bec = m∠aec. the angle addition postulate gives 45° + m∠bec = 90°. so we find m∠bec = 45°. since eb divides ∠aec into two congruent angles, it is the angle bisector.

Explanation:

Step1: Define coordinates

Let E(0,0), A(0,a), C(c,0); angle AEC=90°.

Step2: Angle AEB=45° implies x=y

Vectors EA=(0,a), EB=(x,y). Dot product: ay/(a√(x²+y²))=√2/2 ⇒ x=y.

Step3: Right angle ABC gives y=(a+c)/2

Vectors BA=(-y,a-y), BC=(c-y,-y). Dot product=0 ⇒ 2y=a+c ⇒ y=(a+c)/2.

Step4: Calculate angle BEC

Vectors EB=((a+c)/2,(a+c)/2), EC=(c,0). cos(angle BEC)=1/√2 ⇒ angle BEC=45°.

Step5: Conclude EB bisects ∠AEC

∠AEB=∠BEC=45°, so EB bisects ∠AEC.

Answer:

EB bisects ∠AEC.